if not when is each used?

>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 1
 05012013 07:13

Revision help in partnership with Birmingham City University

 Follow
 2
 05012013 07:25
Someone correct me if I'm wrong, but from what I remember:
λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc
At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E
So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.
I hope that helps. 
agostino981
 Follow
 0 followers
 0 badges
 Send a private message to agostino981
Offline0ReputationRep: Follow
 3
 05012013 07:46
(Original post by ibs2012)
Someone correct me if I'm wrong, but from what I remember:
λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc
At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E
So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.
I hope that helps.
But, what you can do is apply and , where is the frequency and is the wavelength.
EDIT:
(Original post by >>MMM<<)
i am now confused so they are the same or not if not why and when do we use each?!
What I did is to apply the equation for energy level and the equation relating the speed of wave, wavelength and frequency , because it is a photon, you have , hence
If you have something with mass, you can, of course, apply or for relativistic condition. However, you are considering photon, for momentum, you have .Last edited by agostino981; 06012013 at 03:43. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 4
 05012013 07:49
(Original post by agostino981)
Photon is massless, so you can't really apply .
But, what you can do is apply and , where is the frequency and is the wavelength. 
 Follow
 5
 05012013 07:53
(Original post by agostino981)
Photon is massless, so you can't really apply .
But, what you can do is apply and , where is the frequency and is the wavelength. 
 Follow
 6
 05012013 07:59
(Original post by >>MMM<<)
i am now confused so they are the same or not if not why and when do we use each?!
In that case the only use I can find for λ=h/mv is when you are trying to find the de Broglie wavelength of larger objects, ie a car or person moving. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 7
 05012013 08:51
(Original post by ibs2012)
I'm assuming that your variables are m=mass,v=velocity?
In that case the only use I can find for λ=h/mv is when you are trying to find the de Broglie wavelength of larger objects, ie a car or person moving. 
agostino981
 Follow
 0 followers
 0 badges
 Send a private message to agostino981
Offline0ReputationRep: Follow
 8
 05012013 09:27
(Original post by >>MMM<<)
how is that possible? i have read in my book that λ=h/mv can be used in any situation? 
 Follow
 9
 05012013 09:39
(Original post by ibs2012)
Someone correct me if I'm wrong, but from what I remember:
λ=h/p
from mechanics: p=mv
so: λ=h/mv
Photons move at the speed of light, so v=c
λ=h/mc
At the same time, from Einstein: E=mc^2
m=E/c^2
thus: λ=h/[(E/c^2)c]
λ=h/(E/c)
λ=hc/E
So basically both equations are related, but not quite the same. You need to apply them depending on what variables are known and which ones you are trying to find.
I hope that helps.
λ=hc/E is the wavelength of a photon with a certain energy. You usually see the equation in this form: E=hc/λ. and they will usually ask to find the wavelength energy given a certain wavelength.
I dont think they are they same (they might be, but I dont' know). But generally in my course (OCR), λ=h/mv is for particles and E=hc/λ is for photons of light.Last edited by bbcs2k9; 05012013 at 09:41. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 10
 05012013 10:39
(Original post by bbcs2k9)
λ=h/mv is known as the Broglie wavelength and you use it when you are asked to find the associated wavelength of a particle such as an electron with a certain speed.
λ=hc/E is the wavelength of a photon with a certain energy. You usually see the equation in this form: E=hc/λ. and they will usually ask to find the wavelength energy given a certain wavelength.
I dont think they are they same (they might be, but I dont' know). But generally in my course (OCR), λ=h/mv is for particles and E=hc/λ is for photons of light. 
theterminator
 Follow
 0 followers
 1 badge
 Send a private message to theterminator
Offline1ReputationRep: Follow
 11
 05012013 11:04
(Original post by >>MMM<<)
that was the answer i was looking forward to but the problem is that my book says that a beam of moving electrons has associated de broglie wavelength (where at high speed you have to take into account relativistic effects) is approx λ≃hc/E can you please tell me what does that mean?
Last edited by theterminator; 05012013 at 11:06. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 12
 05012013 11:21
(Original post by theterminator)
If you want to calculate the relativistic de broglie wavelength of a particle of rest mass then the equation becomes:

theterminator
 Follow
 0 followers
 1 badge
 Send a private message to theterminator
Offline1ReputationRep: Follow
 13
 05012013 11:37
(Original post by >>MMM<<)
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?! 
 Follow
 14
 05012013 11:51
(Original post by >>MMM<<)
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?! 
agostino981
 Follow
 0 followers
 0 badges
 Send a private message to agostino981
Offline0ReputationRep: Follow
 15
 05012013 11:56
(Original post by >>MMM<<)
thanks but how can we use λ=hc/E to find the de broglie of electrons if it is primarily used for photons (where E is the energy of the electron)?!
If you want to find the deBroglie wavelength of electrons, you need .
If you seek a nonrelativistic result, you apply , if you seek a relativistic result, you apply , where
A pretty obvious reason why you can't use is because electrons cannot reach light speed. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 16
 05012013 12:47
(Original post by agostino981)
Nope, is for photons only.
If you want to find the deBroglie wavelength of electrons, you need .
If you seek a nonrelativistic result, you apply , if you seek a relativistic result, you apply , where
A pretty obvious reason why you can't use is because electrons cannot reach light speed. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 17
 05012013 12:49
(Original post by Pangol)
Look again at bbcs2k9's post above. E=hf (or E=hc/λ, they are the same equation) is for photons. If you wnat to find the deBroglie wavelength of a particle, you need to use λ=h/mv. 
>>MMM<<
 Follow
 0 followers
 2 badges
 Send a private message to >>MMM<<
 Thread Starter
Offline2ReputationRep: Follow
 18
 05012013 12:50
(Original post by theterminator)
I think you may be getting confused with the derivation of the de Broglie equation from above. The above derivation is the one taught but isn't actually the correct way of deriving the de Broglie wavelength. The correct derivation requires more fundamental special relativity but it makes the clear distinction between photons and electrons. Something you should know about the de Broglie wavelength is that it is a relativistic effect, and E=mc^2 is an equation for something that has zero velocity in the frame of reference being used. 
 Follow
 19
 05012013 12:51
(Original post by >>MMM<<)
soo your saying λ=h/mv is can only for particles and λ=hc/E can only for photons and the book has some kind of mistake in it? 
theterminator
 Follow
 0 followers
 1 badge
 Send a private message to theterminator
Offline1ReputationRep: Follow
 20
 05012013 12:58
(Original post by >>MMM<<)
so λ=hc/E can't be used for particles??
Related university courses

Physics with Particle Physics & Cosmology
Swansea University

Physics
University of York

Physics
HeriotWatt University

Physics
University of Hull

Environmental Science and Physics
Keele University

Mathematics and Theoretical Physics
University of St Andrews

Physics with Astrophysics with Professional and Research Placements
University of Bath

Physics with Astrophysics with Science Foundation Year
Keele University

Physics with Nuclear Technology BSc Hons (Sandwich)
University of the West of Scotland

Theoretical Physics (4 years)
University of Birmingham
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 charco
 Mr M
 Changing Skies
 F1's Finest
 rayquaza17
 Notnek
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel